y^'+2y=25sin(x)+40cos(x),y(0)=1

Lösung

y^{^{'}} + 2 y = 25 sin (x) + 40 cos (x), y (0) = 1

y = \frac{5 ( - e ^{2 x} cos ( x ) + 2 e ^{2 x} sin ( x ) + 8 ( \frac{e ^{2 x} s i n ( x )}{5} + \frac{2 e ^{2 x} c o s ( x )}{5} ) )}{e ^{2 x}} - \frac{10}{e ^{2 x}}

Schritte zur Lösung

y^{'} (x) + 2 y = 25 sin (x) + 40 cos (x)

Solve linear ODE:

y = \frac{5 ( - e ^{2 x} cos ( x ) + 2 e ^{2 x} sin ( x ) + 8 ( \frac{e ^{2 x} s i n ( x )}{5} + \frac{2 e ^{2 x} c o s ( x )}{5} ) )}{e ^{2 x}} - \frac{10}{e ^{2 x}}

y = \frac{5 ( - e ^{2 x} cos ( x ) + 2 e ^{2 x} sin ( x ) + 8 ( \frac{e ^{2 x} s i n ( x )}{5} + \frac{2 e ^{2 x} c o s ( x )}{5} ) )}{e ^{2 x}} - \frac{10}{e ^{2 x}}

t an g e n t y = x, (9, 3)

n = 6 \sum \infty \frac{5 ^{n}}{1 2 ^{n}}

3 y^{^{''}} + 17 y^{^{'}} - 6 y = 0

\frac{d}{d x} (\frac{1}{x ^{4}} + \frac{1}{x ^{2}} - x^{3})

\int (x^{2} - 5)^{5} \cdot 2 x d x